Android 引起的:java.lang.NullPointerException [英] Android Caused by: java.lang.NullPointerException

问题描述

我现在编辑了它...我正在检查布局和初始化，但仍然出现相同的错误...是否可能是 Eclipse 错误?

I Edited it now...I'm check the layout and the initialization but still the same error..is it maybe eclipse error?

public class ActivityOwner extends Activity implements OnClickListener {

        EditText ownerUser, ownerPass;
        private Button btnLogin;
        private ProgressDialog pDialog;
        JSONParser jsonParser = new JSONParser();

        private static final String OWNER_LOGIN_URL = "http://192.168.2.5/idot/owner_login.php";

        // ---JSON element ids from response of php script
        private static final String TAG_SUCCESS = "success";
        private static final String TAG_MESSAGE = "message";

        @Override
        protected void onCreate(Bundle savedInstanceState) {
            super.onCreate(savedInstanceState);
            setContentView(R.layout.activity_owner);

            ownerUser = (EditText) findViewById(R.id.owner_login_username);
            ownerPass = (EditText) findViewById(R.id.owner_login_password);

            TextView tvLinkToRegister = (TextView) findViewById(R.id.link_to_register);
            tvLinkToRegister.setOnClickListener(new View.OnClickListener() {

                @Override
                public void onClick(View view) {
                    // TODO Auto-generated method stub
                    Intent registrationForm = new Intent(ActivityOwner.this,
                            ActivityOwnerRegister.class);
                    startActivity(registrationForm);
                }
            });

            btnLogin = (Button) findViewById(R.id.btnOwnerLogin);
            btnLogin.setOnClickListener(this);
        }

        @Override
        public void onClick(View v) {
            new AttemptLogin().execute();
        }

        class AttemptLogin extends AsyncTask<String, String, String> {

            @Override
            protected void onPreExecute() {
                super.onPreExecute();
                pDialog = new ProgressDialog(ActivityOwner.this);
                pDialog.setMessage("Attempting login...");
                pDialog.setIndeterminate(false);
                pDialog.setCancelable(true);
                pDialog.show();
            }

            @Override
            protected String doInBackground(String... args) {

                // Check for success tag
                int success;
                String username = ownerUser.getText().toString();
                String password = ownerPass.getText().toString();

                try {
                    // Building Parameters
                    List<NameValuePair> params = new ArrayList<NameValuePair>();
                    params.add(new BasicNameValuePair("owner_username", username));
                    params.add(new BasicNameValuePair("owner_password", password));

                    Log.d("request!", "starting");
                    // getting product details by making HTTP request
                    JSONObject json = jsonParser.makeHttpRequest(OWNER_LOGIN_URL,
                            "POST", params);

                    // check your log for json response
                    Log.d("Login attempt", json.toString());

                    // json success tag
                    success = json.getInt(TAG_SUCCESS);
                    if (success == 1) {
                        Log.d("Login Successful!", json.toString());
                        // save user data
                        SharedPreferences sp = PreferenceManager
                                .getDefaultSharedPreferences(ActivityOwner.this);
                        Editor edit = sp.edit();
                        edit.putString("owner_username", username);
                        edit.commit();

                        Intent i = new Intent(ActivityOwner.this,
                                ActivityOwnerSuccessLogin.class);
                        pDialog.dismiss();
                        finish();
                        startActivity(i);

                        return json.getString(TAG_MESSAGE);
                    } else {
                        Log.d("Login Failure!", json.getString(TAG_MESSAGE));
                        return json.getString(TAG_MESSAGE);
                    }
                } catch (JSONException e) {
                    e.printStackTrace();
                }
                return null;
            }

            @Override
            protected void onPostExecute(String result) {
                // dismiss the dialog once product deleted
                pDialog.dismiss();
                if (result != null) {
                    Toast.makeText(ActivityOwner.this, result, Toast.LENGTH_SHORT)
                            .show();
                }
            }

        }
    }

------LOGCAT-------

------LOGCAT-------

11-14 21:30:52.517: E/AndroidRuntime(18929): Caused by: java.lang.NullPointerException
11-14 21:30:52.517: E/AndroidRuntime(18929):    at ph.idot.owner.ActivityOwner$AttemptLogin.doInBackground(ActivityOwner.java:88)

推荐答案

A.B.C 形式的一行上的空指针异常意味着点左侧的某些内容为空.在这个例子中:

A null pointer exception on a line of the form A.B.C means that something on the left-hand side of a dot is null. In this example :

• 要么A为空；
• 或 A.B 为空.

没有其他原因.因此，由于 textView.getText() 从不返回 null，唯一的可能性是 user 为 null.

There can be no other reason. So, as textView.getText() never returns null, the only possibility is that user is null.

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